An object's two dimensional velocity is given by $v (t) = (t^{2} - 2 t, cos π t - t)$ $v(t) = ( t^2 - 2t , cospit - t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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An object's two dimensional velocity is given by $v (t) = (t^{2} - 2 t, cos π t - t)$ $v(t) = ( t^2 - 2t , cospit - t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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Answer 1 · 2017-05-06T08:35:55

$answer : a (1) = (0, - 1)$ $"answer :"a(1)=(0,-1)$

Explanation:

$\frac{d}{d t} v (t) = (\frac{d}{d t} (t^{2} - 2 t), \frac{d}{d t} (cos π t - t))$ $d/(d t) v(t)=(color(red)(d/(d t)(t^2-2t)),color(green)(d/(d t)(cos pi t-t)))$

$\frac{d}{d t} v (t) = a (t)$ $d/(d t) v(t)=a(t)$

$a (t) = (2 t - 2, - π sin π t - 1)$ $a(t)=(color(red)(2t-2),color(green)(-pi sin pi t-1))$

$solve t=1$ $"solve t=1"$

$a (1) = (2 \cdot 1 - 2, - π sin π \cdot 1 - 1)$ $a(1)=(2*1-2,-pi sin pi *1-1)$

$a (1) = (0, - π sin π - 1)$ $a(1)=(0,-pi sin pi-1)$

$sin π = 0$ $sin pi=0$

$a (1) = (0, - 1)$ $a(1)=(0,-1)$

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An object's two dimensional velocity is given by $v (t) = (t^{2} - 2 t, cos π t - t)$ $v(t) = ( t^2 - 2t , cospit - t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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An object's two dimensional velocity is given by v(t) = ( t^2 - 2t , cospit - t )v(t)=(t2−2t,cosπt−t)v(t) = ( t^2 - 2t , cospit - t ). What is the object's rate and direction of acceleration at t=1 t=1t=1 ?

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An object's two dimensional velocity is given by $v (t) = (t^{2} - 2 t, cos π t - t)$ $v(t) = ( t^2 - 2t , cospit - t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?